Question

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CuSO}_{4}(a q)\) b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{MgI}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\mathrm{AlBr}_{3}(a q)\)

Short answer

Precipitates will form in the following cases: a. \(\mathrm{Hg}_{2}\mathrm{S}\mathrm{O}_{4}\) c. \(\mathrm{MgCO}_3\) d. \(\mathrm{Al}_{2}\left(\mathrm{CrO}_{4}\right)_{3}\)

Step-by-step solution

Case A: Mixing Hg2(NO3)2 and CuSO4

In this case, we need to consider the possible products of the mixing: \(\mathrm{Hg}_{2}\mathrm{S}\mathrm{O}_{4}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\). Checking the solubility rules, we find out that \(\mathrm{Hg}_{2}\mathrm{S}\mathrm{O}_{4}\) is insoluble and will form a precipitate. Meanwhile, \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is soluble and will not form a precipitate.

Case B: Mixing Ni(NO3)2 and CaCl2

In this case, the possible products of the mixing are \(\mathrm{NiCl}_{2}\) and \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\). According to the solubility rules, both \(\mathrm{NiCl}_{2}\) and \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) are soluble in water. Therefore, no precipitate will form in this case.

Case C: Mixing K2CO3 and MgI2

In this case, the possible products of the mixing are \(\mathrm{K}\mathrm{I}\) and \(\mathrm{MgCO}_3\). According to the solubility rules, \(\mathrm{K}\mathrm{I}\) is soluble, but \(\mathrm{MgCO}_3\) is insoluble and will form a precipitate.

Case D: Mixing Na2CrO4 and AlBr3

In this case, the possible products of the mixing are \(\mathrm{NaBr}\) and \(\mathrm{Al}_{2}\left(\mathrm{CrO}_{4}\right)_{3}\). Based on the solubility rules, \(\mathrm{NaBr}\) is soluble, but \(\mathrm{Al}_{2}\left(\mathrm{CrO}_{4}\right)_{3}\) is insoluble and will form a precipitate. In summary, precipitates will form in the following cases: a. \(\mathrm{Hg}_{2}\mathrm{S}\mathrm{O}_{4}\) c. \(\mathrm{MgCO}_3\) d. \(\mathrm{Al}_{2}\left(\mathrm{CrO}_{4}\right)_{3}\)

Key concepts

Solubility Rules

The solubility rules are guidelines that help us predict whether a compound will dissolve in water or form a precipitate. These rules are based on experimental data and provide insight into the behavior of ionic compounds in an aqueous solution. Knowing these rules helps chemists determine the solubility of salts and their tendency to precipitate.

• Compounds containing alkali metal ions (like Na\(^+\), K\(^+\)) and the ammonium ion (NH\(_4^+\)) are always soluble.
• Most nitrates (NO\(_3^-\)) and acetates (CH\(_3\)COO\(^-\)) are soluble.
• Chlorides (Cl\(^-\)), bromides (Br\(^-\)), and iodides (I\(^-\)) are generally soluble, except for those of silver (Ag\(^+\)), lead (Pb\(^{2+}\)), and mercury (Hg\(^{2+}\)).
• Sulfates (SO\(_4^{2-}\)) are soluble except for those of barium (Ba\(^{2+}\)), calcium (Ca\(^{2+}\)), and lead (Pb\(^{2+}\)).
• Carbonates (CO\(_3^{2-}\)) and phosphates (PO\(_4^{3-}\)) are generally insoluble except when paired with alkali metals or NH\(_4^+\).
  

This exercise demonstrates the application of solubility rules by predicting the formation of precipitates when two aqueous solutions are mixed. For example, in case (a), the formation of Hg\(_2\)SO\(_4\) as a precipitate occurs because it is insoluble according to these rules.

Precipitation Reactions

Precipitation reactions occur when two soluble ionic compounds in aqueous solutions are mixed, resulting in the formation of an insoluble compound known as a precipitate. This process is fundamental in chemistry for isolating and identifying compounds.
When the ions in two compounds come together to form a compound that does not dissolve in water, a solid is formed. This is what we call a precipitate. For example, when mixing \(\text{K}_2\text{CO}_3\) and \(\text{MgI}_2\), \(\text{MgCO}_3\) forms as a precipitate because it is insoluble in water.

• Ingredients: The reactants are two aqueous solutions of ionic compounds.
• Mixing: The reactants are mixed, resulting in the exchange of ions.
• Outcome: A new ionic compound forms as an insoluble solid which then precipitates out of the solution.
• Separation: This precipitate can be separated by filtration, allowing for further analysis or practical use.
  

In summary, precipitation reactions play a key role in chemical analysis and synthesis by creating insoluble compounds from aquatic ion exchanges.

Ionic Equations

Ionic equations are an essential tool in chemistry, allowing us to focus on the species that actually change during a reaction. These equations break down soluble ionic compounds into their constituent ions. Understanding these helps students grasp how elements in ionic compounds interact.
There are two main types of ionic equations:

• Complete Ionic Equations: These show all the ions present in the reaction. For example, mixing Na\(_2\)CrO\(_4(aq)\) with AlBr\(_3(aq)\) involves breaking these compounds into Na\(^+\), CrO\(_4^{2-}\), Al\(^{3+}\), and Br\(^-\) ions.
• Net Ionic Equations: These only include the ions and molecules directly involved in the formation of the precipitate, eliminating spectator ions. This means only the ions that form the solid precipitate, such as Mg\(^{2+}\) and CO\(_3^{2-}\) from MgCO\(_3\), are shown.

Ionic equations help simplify complex reactions by stripping down the reaction to its essence: the formation of the precipitate. They are crucial in understanding which ions interact to form new compounds in precipitation reactions.